Question
Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant
The ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Answer
Ionic product,
Kw = [H+][OH-]
Let [H+] = x.
Since [H+] = [OH-], Kw = x2
⇒ Kw at 310K is 2.7 × 10-14
∴ 2.7 ×10-14 = x2
⇒ x = 1.64 × 10-7
⇒ [H+] = 1.64 × 10-7 ⇒ pH = -log [H+] = -log [1.64 × 10-7] = 6.78
Hence, the pH of neutral water is 6.78.